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于 2018-07-21 21:18:11 发布

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分类专栏：多校联合题目

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本文介绍了一个有趣的算法问题，即通过输入公司员工间的上下级关系，计算具有特定数量下属的管理者人数。文章提供了一种解决方案，并附带了实现该算法的C++代码。

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Problem Description

It’s an interesting experience to move from ICPC to work, end my college life and start a brand new journey in company.
As is known to all, every stuff in a company has a title, everyone except the boss has a direct leader, and all the relationship forms a tree. If A’s title is higher than B(A is the direct or indirect leader of B), we call it A manages B.
Now, give you the relation of a company, can you calculate how many people manage k people.

Input

There are multiple test cases. Each test case begins with two integers n and k, n indicates the number of stuff of the company. Each of the following n-1 lines has two integers A and B, means A is the direct leader of B. 1 <= n <= 100 , 0 <= k < n 1 <= A, B <= n

Output

For each test case, output the answer as described above.

Sample Inpu

7 2 1 2 1 3 2 4 2 5 3 6 3 7

Sample Output

Author

ZSTU

Source

2015 Multi-University Training Contest 3

题目大概：

给出n个人和（n-1）个关系，a b表示a 是 b的领导。然后，求有k个下属的人的数量。

思路：

本来想用并查集，然后随便改了改，改的也不是路径压缩了，交了一发，过了，看来这份题的出题人比较良心。。。

代码：

#include <bits/stdc++.h>

using namespace std;
const int maxn=110;
int fa[maxn],shu[maxn];
struct poin
{
    int u,v;
}G[maxn];
int cmp(poin a,poin b)
{
    return a.u>b.u;
}
int f(int x)
{
    if(fa[x]==-1)return x;
    else return fa[x]=f(fa[x]);
}
void bond(int x,int y)
{
    int fx=f(x);
    int fy=f(y);
    if(fx!=fy)
    {
        fa[fy]=fx;
        shu[x]+=shu[y]+1;
        //cout<<x<<" "<<y<<" "<<shu[fx]<<endl;
    }
}
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        memset(fa,-1,sizeof(fa));
        for(int i=1;i<=n;i++)shu[i]=0;
        for(int i=1;i<n;i++)
        {
            scanf("%d%d",&G[i].u,&G[i].v);
        }
        sort(G+1,G+n,cmp);
        for(int i=1;i<n;i++)
        {
            bond(G[i].u,G[i].v);
        }
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            if(shu[i]==k)ans++;
            //cout<<shu[i]<<endl;
        }
        printf("%d\n",ans);
    }
    return 0;
}